Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Given:

$r_d = 40k\Omega$

$kn =0.4×10^{-3}$

$V_{GS(th)}=3V$

$Av=-g_m(r_d|| R_D)$

Rin=R1||R2

$Ro=r_d||R_D$

We know that,

$g_m= 2kn(V_{GSQ}-V_T)$

$g_m= 2×0.4×10^{-3}(V_{GSQ}-3)$……………..(1)

To find $g_m and V_{GSQ}$ we have to perform DC analysis, all the coupling capacitor acts as open circuit, as shown in Figure 2:

$I_{DS}=kn(V_{GS}-V_T)^2$

$I_{DS}=0.4\times10^{-3}(V_{GS}-3)^2$----------------(2)

But voltage divider biasing equation …